Estimates of $\Delta|\nabla u|$ for harmonic function $u$

Question

The well-known Bochner formula and refined Kato inequality (for harmonic function) tells us that for a harmonic function $u$ in $B_{2}\subset\mathbb{R}^{n}$, $$ \frac{1}{n}|\nabla^{2}u|^{2}\leqslant |\nabla^{2}u|^{2}-|\nabla|\nabla u||^{2}= |\nabla u|\Delta|\nabla u|\leqslant |\nabla^{2}u|^{2}. $$ It seems that there is a certain connection between $\Delta|\nabla u|$ and $|\nabla^{2}u|^{2}$. My question is how to obtain an analogous $L^{p}$ estimate for $\Delta|\nabla u|$. Let me fully state my question below.

For any $\delta>0$, does there exist $\epsilon>0$ such that if $u$ is a harmonic function in $B_{2}\subset\mathbb{R}^{n}$ with $$ \begin{split} &|\nabla u|\leqslant 1, \\ &\int_{B_{2}}||\nabla u|-1| < \epsilon\\ &\int_{B_{2}}|\nabla^{2}u|^{2p} < \epsilon \end{split} $$ for some integer $p\geqslant1$, then $$ \int_{B_{1}}(\Delta|\nabla u|)^{p}<\delta\; ? $$ Note that we only need the estimate in a smaller ball, so one may use a cutoff function. And it is easy when $p=1$, since we can multiply a good cutoff function with bounded laplacian and then integrate by parts. But for integer $p\geqslant2$, I can’t see how to do it.

Answer 1

Yes, this inequality is true, and in fact the assumption about the smallness of $D^2u$ in $L^{2p}$ is not needed.

Using that $|\nabla u|$ is $\epsilon$-close to one in $L^1(B_2)$, we can find a point in $B_1$ where $\nabla u$ is $C(n)\epsilon$-close to the unit sphere, say $-e_1$ after a rotation. Then $u_1 + 1$ is nonnegative (since $|\nabla u| \leq 1$), harmonic, and $C(n)\epsilon$-close to zero somewhere in $B_1$. By the Harnack inequality, $u_1$ is pointwise $C(n)\epsilon$-close to $-1$ in $B_{3/2}$. Using again that $|\nabla u| \leq 1$ we conclude that $\nabla u$ is pointwise $C(n)\sqrt{\epsilon}$-close to $-e_1$ in $B_{3/2}$. The interior derivative estimates for harmonic functions then imply that $|D^ku|$ are pointwise $C(n,k)\sqrt{\epsilon}$-small in $B_1$ for $k \geq 2$. This completes the proof with $\delta = C(n,p)\epsilon^{p}$.